New York Journal of Mathematics
New York J. Math. 11 (2005) 649–658.
The graph traces of finite graphs and applications
to tracial states of C ∗ -algebras
Matthew Johnson
Abstract. We determine the extreme points of the set of graph traces of
norm one for any finite graph E satisfying Condition (K). We also describe an
application to the space of tracial states on the graph C ∗ -algebra.
Contents
1. Introduction
2. Preliminaries
3. Finite graphs with no loops
4. Finite graphs with loops
5. An example
References
649
650
651
655
656
658
1. Introduction
If E is a directed graph, the graph algebra C ∗ (E) is the universal C ∗ -algebra
generated by a collection of partial isometries satisfying certain relations determined
by E. This paper will focus on the space of tracial states on C ∗ (E), denoted
T (C ∗ (E)), which is the set of all positive, linear functionals τ : C ∗ (E) → C of norm
one such that τ (ab) = τ (ba) for all a, b ∈ C ∗ (E). If E is a finite graph satisfying
the so-called Condition (K), to be defined below, then T (C ∗ (E)) is isomorphic to
the set of graph traces of norm one defined on E, denoted T (E) [5]. Since T (E)
is a compact convex set in a certain locally convex topological vector space, the
Krein–Millman Theorem implies that T (E) is the closed convex hull of its extreme
points [4]. The objective of this paper is to provide an effective way of computing the
extreme points of T (E) so that T (E) and subsequently T (C ∗ (E)) will be completely
and effectively described when E is a finite graph satisfying Condition (K).
Received December 7, 2004, and in revised form, December 4, 2005.
Mathematics Subject Classification. 46L55.
Key words and phrases. C ∗ -algebras, directed graph, trace, tracial state, convex set.
This research was supported by NSF Grant DMS-0070405 and NSF Postdoctoral Fellowship
DMS-0201960.
ISSN 1076-9803/05
649
650
Matthew Johnson
In more detail, we show that the set extreme points of T (E), where E is a
finite graph satisfying Condition (K), consists of a finite number of graph traces
corresponding to certain sinks of the graph. These graph traces are easy to compute
in general, and an example is provided to show how to find them. As a result, one
can use these graph traces to determine the space of tracial states on C ∗ (E).
Acknowledgements. This work is the result of a research project undertaken
while the author was an undergraduate at the University of Iowa. The author
thanks his mentor, Mark Tomforde, for his help and guidance throughout the
project.
2. Preliminaries
A (directed) graph E = (E 0 , E 1 , r, s) consists of a countable set E 0 of vertices,
a countable set E 1 of edges, and maps r, s : E 1 → E 0 identifying the range and
source of each edge. While our focus will be on finite graphs, i.e., graphs where the
sets of vertices and edges are finite, we shall make this assumption only when it is
relevant to the proof of our main result, Theorem 4.4.
A vertex v ∈ E 0 is called a sink if |s−1 (v)| = 0, and v is called an infinite emitter
if |s−1 (v)| = ∞. A singular vertex is a vertex which is either a sink or an infinite
emitter. We will let SE ⊆ E 0 denote the set of all of the sinks of E.
A path is a finite sequence of edges α = e1 . . . en with r(ei ) = s(ei+1 ) for 1 ≤ i ≤
n − 1; we let |α| = n denote
the path’s length. Let E n be the set of all paths in E
∞
∗
of length n and let E := n=0 E n . We extend r and s to E ∗ in the natural way:
s(e1 . . . en ) = s(e1 ) and r(e1 . . . en ) = r(en ).
If v and w are vertices, we say that v ≥ w if there is a path e1 . . . en with
s(e1 ) = v and r(en ) = w. Let n(w, v) := #{α ∈ E ∗ : s(α) = w and r(α) = v} and
n(v) := #{α ∈ E ∗ : r(α) = v}. A loop is a path e1 . . . en with s(e1 ) = r(en ) and
we call s(e1 ) the base point of the loop. A loop e1 . . . en is simple if s(ei ) = s(e1 )
for all i ∈ {2, 3, . . . , n}.
Definition 2.1. A graph E satisfies Condition (K) if no vertex is the base point
of exactly one simple loop.
Definition 2.2. If E is a graph, then a graph trace on E is a function g : E 0 →
[0, ∞) with the following two properties:
(1) For any nonsingular vertex v ∈ E 0 we have g(v) = {e∈E 1 :s(e)=v} g(r(e)).
(2) For any infinite emitter
v ∈ E 0 and any finite collection of edges e1 , . . . , en ∈
n
−1
s (v) we have g(v) ≥ i=1 g(r(ei )).
We define the norm of a graph trace g to be the (possibly infinite) value ||g|| :=
v∈E 0 g(v). Let T (E) denote the set of all graph traces of norm one. We will
0
view T (E) as a subset of the space CE consisting of all C-valued functions on
0
E , endowed with the product topology. This space has the subbasis {Nv, (g) :
v ∈ E 0 , > 0, and g ∈ T (E)}, where Nv, (g) := {h ∈ T (E) : |h(v) − g(v)| < }.
0
Evidently, T (E) is a convex subset of CE , and it is easy to see that T (E) is compact
in this topology. Indeed, T (E) is a closed subset of D
0
disc in C), which is a compact subset of CE .
E0
(where D is the closed unit
Graph traces and C ∗ -algebras
651
Recall that an extreme point of a convex set X in a vector space is an element
x ∈ X such that if x = tx1 +(1−t)x2 for some points x1 and x2 in X and for some t,
0 < t < 1, then either x = x1 or x = x2 . As is customary, we write ext(X) to denote
the set of all of the extreme points of X. From the Krein–Millman Theorem [4, pg.
70] we know that since T (E) is a compact convex subset of a locally convex space it
is the closed convex hull of its extreme points. Thus, if we know the extreme points
of T (E), then we know T (E), especially when E is finite, since then any graph
trace in T (E) can be written as a convex combination of extreme points. Indeed,
by Carathéodory’s theorem, each point may be written as a convex combination of
|E 0 | + 1 or fewer extreme points in T (E).
For a graph E, a Cuntz–Krieger E-family is a set of mutually orthogonal projections {pv : v ∈ E 0 } and a set of partial isometries {se : e ∈ E 1 } with orthogonal
ranges that satisfy the Cuntz–Krieger relations:
(1)
s∗e se = pr(e) for every e ∈ E 1 ;
(2)
se s∗e ≤ ps(e) for every e ∈ E 1 ;
pv =
se s∗e for every v ∈ E 0 that is not a singular vertex.
(3)
{e:s(e)=v}
It is possible to associate a C ∗ -algebra with E; namely the C ∗ -algebra generated
by a universal Cuntz–Krieger E-family, C ∗ (E) := {pv , se : v ∈ E 0 and e ∈ E 1 } [2,
§1].
We have a map ι : T (C ∗ (E)) → T (E) defined by
ι(τ )(v) = τ (pv )
where v ∈ E 0 and τ ∈ T (C ∗ (E)).
It is shown in [5] that if E satisfies Condition (K) then ι is an isomorphism. In
order to calculate T (C ∗ (E)) we use ι to map the extreme points of T (E) to the
extreme points of T (C ∗ (E)).
In this paper it will be shown that the extreme points of T (E) consist of graph
traces of the following type.
Definition 2.3. For a finite graph E with no loops and v ∈ SE , define gv : E 0 →
[0, ∞) by
n(w, v)
for w ∈ E 0 .
gv (w) :=
n(v)
Note that if E is finite with no loops, then the term n(w, v) is always finite and
nonnegative and the term n(v) is always finite and positive.
3. Finite graphs with no loops
Lemma 3.1. If v, w ∈ E 0 with v ≥ w, then g(v) ≥ g(w) for any graph trace g on
E.
Proof. If v ≥ w then there is a path e1 . . . en with s(e1 ) = v and r(en ) = w. Since
g is a graph trace it takes nonnegative values and thus
g(v) =
g(r(e)) ≥ g(r(e1 )) =
g(r(e)) ≥ · · · ≥ g(r(en )) = g(w).
s(e)=v
s(e)=r(e1 )
Lemma 3.2. If e1 . . . en is a loop and g is a graph trace on E, then g(s(e1 )) =
g(s(ei )) for all 1 ≤ i ≤ n.
652
Matthew Johnson
Proof. Since e1 . . . en is a loop we know that s(e1 ) = r(en ) and that
s(e1 ) ≥ s(e2 ) ≥ · · · ≥ s(en ) ≥ r(en ) = s(e1 ).
Using Lemma 3.1 we have
g(s(e1 )) ≥ g(s(e2 )) ≥ · · · ≥ g(s(en )) ≥ g(s(e1 ))
from which it follows that g(s(e1 )) = g(s(ei )) for all 1 ≤ i ≤ n.
Proposition 3.3. Let E = (E 0 , E 1 , r, s) be a graph that satisfies Condition (K),
let g be a graph trace on E, and let e1 . . . en be a loop in E. Then g(s(ei )) = 0 for
all 1 ≤ i ≤ n.
Proof. Let v = s(ei ) for some 1 ≤ i ≤ n. Since E satisfies Condition (K) and
v is the base point of a loop, we know that v is the base point of two distinct
loops f1 . . . fm and h1 . . . hl . Because these two loops are distinct, there exists
1 ≤ j ≤ min{m, l} for which s(fj ) = s(hj ) but fj = hj .
Thus, using the definition of a graph trace and Lemma 3.2 we have
g(v) = g(s(fj ))
=
(by Lemma 3.2)
g(r(e))
(by Definition 2.2)
{e∈E 1 :s(e)=s(fj )}
≥ g(r(fj )) + g(r(hj ))
= 2g(v)
(by Lemma 3.2).
Since g has nonnegative values, this implies that g(v) = 0.
Lemma 3.4. Let E be a finite graph with no loops and let SE ⊆ E 0 be the set of
all sinks of E. If g1 and g2 are graph traces on E and g1 (v) = g2 (v) for all v ∈ SE ,
then g1 = g2 .
Proof. Assume that g1 = g2 and g1 (v) = g2 (v) for all v ∈ SE . Since g1 = g2 there
exists w ∈ E 0 such that
g1 (w) = g2 (w).
we know that w ∈
/ SE . By the definition of
Because g1 (v) = g2 (v) for all v ∈ SE a graph trace we know that gi (w) = {e∈E 1 :s(e)=w} gi (r(e)) for i = 1, 2. Hence,
there exists e ∈ E 1 with s(e) = w and g1 (r(e)) = g2 (r(e)). Because E is a finite
graph with no loops we are able to repeat this process and eventually end at a sink
v0 with w ≥ v0 and
g1 (v0 ) = g2 (v0 ).
Thus, it is not the case that g1 (v) = g2 (v) for all v ∈ SE and we arrive at a
contradiction.
Lemma 3.5. Let E be a finite graph with no loops. If v ∈ SE , then gv ∈ T (E).
Proof. If w ∈ E 0 and w = v then n(w, v) = {e∈E 1 :s(e)=w} n(r(e), v) since any
path α from w to v must be of the form eβ for a unique e ∈ s−1 (w) and a unique
path β. Thus
n(r(e), v)
n(w, v)
.
=
n(v)
n(v)
s(e)=w
Graph traces and C ∗ -algebras
From which it follows that
gv (w) =
653
gv (r(e))
s(e)=w
and hence
gv is a graph trace on E. In order to show gv ∈ T (E) we first note that
n(v) = w∈E 0 n(w, v). Dividing both sides by n(v) gives
1=
n(w, v)
gv (w).
=
n(v)
0
0
w∈E
w∈E
Therefore gv ∈ T (E) for all v ∈ SE .
Lemma 3.6. Let E be a finite graph with no loops. If g is a graph trace on E,
then g(w) ≥ n(w, v)g(v) for all v, w ∈ E 0 .
Proof. If there is no path from w to v then n(w, v) = 0 and the claim holds
trivially. Thus we need only consider the case were w ≥ v.
Let w, v ∈ E 0 with w ≥ v. This will be a proof by induction on k, where
k := max {|α| : α ∈ E ∗ with s(α) = w and r(α) = v }.
If k = 0, then w = v and n(w, v) = n(v, v) = 1, and the lemma holds. For the
inductive step, assume that the lemma holds for some particular k ≥ 0. Choose
w ∈ E 0 such that
max {|α| : α ∈ E ∗ with s(α) = w and r(α) = v} = k + 1.
Let α0 ∈ E ∗ with s(α0 ) = w, r(α0 ) = v, and |α0 | = k + 1. Write α0 = e0 β0 for
e0 ∈ E 1 and β0 ∈ E ∗ . Note that |β0 | = k.
Observe that
max {|α| : α ∈ E ∗ with s(α) = r(e0 ) and r(α) = v } = |β0 | = k.
By the inductive hypothesis we know that g(r(e)) ≥ n(r(e), v)g(v). Using the
definition of a graph trace it follows that
g(r(e)) ≥
n(r(e), v)g(v) = g(v)
n(r(e), v) ≥ g(v)n(w, v).
g(w) =
s(e)=w
s(e)=w
s(e)=w
0
By induction g(w) ≥ n(w, v)g(v) for all v, w ∈ E .
Theorem 3.7. If E is a finite graph with no loops, then
ext(T (E)) = {gv ∈ T (E) : v ∈ SE }.
Proof. Suppose that v ∈ SE and that gv ∈ T (E) is a graph trace such that
gv = tg1 + (1 − t)g2 for 0 < t < 1 and g1 , g2 ∈ T (E). In order to show that gv is an
extreme point we need only show that either gv = g1 or gv = g2 . By Lemma 3.4 it
suffices to show that the two graph traces agree at all of the sinks of E.
We first evaluate gv = tg1 + (1 − t)g2 at v0 ∈ SE where v0 = v. We know that
since both v0 and v are in SE there is no path from v0 to v, and hence gv (v0 ) = 0.
So we now have
gv (v0 ) = 0 = tg1 (v0 ) + (1 − t)g2 (v0 ).
Since both g1 and g2 are graph traces it follows that both g1 (v0 ) and g2 (v0 ) are
nonnegative. Because 0 < t < 1 it is clear that g1 (v0 ) = 0 and g2 (v0 ) = 0. Thus,
gv (v0 ) = g1 (v0 ) = g2 (v0 ) for all v0 ∈ SE with v0 = v.
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Matthew Johnson
Since gi ∈ T (E) for i = 1, 2 we know from Lemma 3.6 that
gi (w) ≥
gi (w) ≥
n(w, v)gi (v) = n(v)gi (v),
1=
w∈E 0
{w∈E 0 :w≥v}
{w∈E 0 :w≥v}
from which it follows that
1
≥ gi (v).
n(v)
We shall now evaluate gv = tg1 + (1 − t)g2 at v. Using the definition of gv we have
1
= gv (v)
n(v)
= tg1 (v) + (1 − t)g2 (v)
1
≤ tg1 (v) + (1 − t)
n(v)
1
1
−t
= tg1 (v) +
.
n(v)
n(v)
Subtracting
1
n(v)
from both sides yields
0 ≤ tg1 (v) − t
1
n(v)
1
≤ g1 (v).
n(v)
So we now have
1
1
≤ g1 (v) ≤
.
n(v)
n(v)
1
From this we conclude that g1 (v) = n(v)
= gv (v). Therefore by Lemma 3.4, g1 = gv
because g1 and gv agree at each sink. Hence
{gv ∈ T (E) : v ∈ SE } ⊆ ext(T (E)).
To see the reverse inclusion, let g ∈ T (E) and set tv = n(v)g(v) for v ∈ SE .
Then for any w ∈ SE we have
tv gv (w) =
tv gv (w)
v∈SE
v∈SE
= tw gw (w)
1
= tw
n(w)
= g(w).
Thus by Lemma 3.4 we have g =
v∈SE tv gv and any element of T (E) can be
written as a convex combination of the gv ’s. Hence
ext(T (E)) = {gv ∈ T (E) : v ∈ SE }.
Graph traces and C ∗ -algebras
655
4. Finite graphs with loops
In this section we extend the results of Section 3 to all finite graphs satisfying
Condition (K).
If E is a (not necessarily finite) graph, then a subset H ⊆ E 0 is called hereditary
if s(e) ∈ H implies that r(e) ∈ H. If H ⊆ E 0 is hereditary, H is called saturated if
whenever v ∈ E 0 with 0 < |s−1 (v)| < ∞, then r(s−1 (v)) ⊆ H implies that v ∈ H.
For a finite graph E, let H := {v ∈ E 0 : s(α) ≥ v for some loop α ∈ E ∗ }. It
is clear that H is hereditary since if s(e) ∈ H for some e ∈ E 1 then there exists a
loop α such that s(α) ≥ s(e). Thus by definition we know that s(e) ≥ r(e), and by
transitivity we know that s(α) ≥ r(e) which implies that r(e) ∈ H.
Let H0 = H and
Hn = Hn−1 ∪ {v ∈ E 0 : 0 < |s−1 (v)| < ∞ and r(s−1 (v)) ⊆ Hn−1 }.
∞
The set H = n=0 Hn is called the saturation of H, and it is the smallest saturated
hereditary set containing H. Let E\H be the graph defined by
E\H := E 0 \H, E 1 \r−1 (H), r|E 0 \H , s|E 0 \H .
Lemma 4.1. Let E be a finite graph which satisfies Condition (K) and let g be a
graph trace on E. Define the hereditary set
H := v ∈ E 0 : s(α) ≥ v for some loop α ∈ E ∗ .
If H is the saturation of H as defined above, then g(w) = 0 for all w ∈ H.
∞
Proof. Since H = n=0 Hn it suffices to show that if w ∈ Hn then g(w) = 0 for
all nonnegative integers n. We shall prove this by induction on n.
If w ∈ H0 then by definition there is a loop α ∈ E ∗ such that s(α) ≥ w. By
Lemma 3.1 we know that g(s(α)) ≥ g(w). But g(s(α)) = 0 by Proposition 3.3 and
it follows that g(w) = 0.
Assume that g(w) = 0 for all w ∈ Hn . Let w ∈ Hn+1 ; it follows that either
w ∈ Hn or w ∈ {v ∈ E 0 : 0 < |s−1 (v)| < ∞ and r(s−1 (v)) ⊆ Hn }. If w ∈ Hn
then by the inductive hypothesis g(w) = 0. If w ∈ {v ∈ E 0 : 0 < |s−1 (v)| <
∞ and r(s−1 (v)) ⊆ Hn } then r(s−1 (w)) ⊆ Hn , from which it follows that that
g(r(e)) = 0 for all e ∈ s−1 (w) ⊆ Hn . By the definition of a graph trace we have
g(r(e)) = 0.
g(w) =
{e∈E 1 :s(e)=w}
Therefore, g(w) = 0 for all w ∈ Hn+1 . By induction we conclude that if w ∈ Hn
then g(w) = 0 for all nonnegative integers n.
Definition 4.2. If E is a graph satisfying Condition (K) and v ∈ SE\H , define
gv ∈ T (E) by
gv (w) w ∈ E\H
gv (w) =
0
w∈H
where gv ∈ T (E\H) is the graph trace of Definition 2.3. (Note that gv is a graph
trace since H is saturated and hereditary, and ||
gv || = ||gv || = 1 so gv ∈ T (E).)
We now prove that T (E) may be identified with T (E\H).
656
Matthew Johnson
Lemma 4.3. Let E be a finite graph satisfying Condition (K). Define the hereditary set H := {v ∈ E 0 : s(α) ≥ v for some loop α ∈ E ∗ }. If H denotes its
saturation, then φ : T (E) → T (E\H) defined by φ(g) = g|E 0 \H is a homeomorphism.
Proof. We shall first show that φ is an injective function. For g1 , g2 ∈ T (E) assume
that φ(g1 ) = φ(g2 ). It follows that g1 and g2 agree on E 0 \H. By Lemma 4.1 we
know that any graph trace in T (E) will be zero on H. Thus, g1 and g2 agree on all
of E and by definition g1 = g2 . Therefore, φ is injective.
Given g ∈ T (E\H) define g ∈ T (E) by
g(v) v ∈ E\H
g(v) =
0
v ∈ H.
It is clear that g is a graph trace and ||
g || = ||g|| = 1, so g ∈ T (E). Also φ(
g) = g
so that φ is surjective. Since a continuous bijection between compact Hausdorff
spaces is a homeomorphism, the result follows.
Theorem 4.4. If E is a finite graph satisfying Condition (K) then
ext(T (E)) = gv ∈ T (E) : v ∈ SE\H .
Proof. By Lemma 4.3 we have that φ : T (E) → T (E\H) is a homeomorphism.
Thus, ext(T (E)) = φ−1 (ext(T (E\H))). However,
ext(T (E\H) = gv ∈ T (E) : v ∈ SE\H
by Theorem 3.7. Hence
φ−1 (ext(E\H))) = φ−1 (gv ) : v ∈ SE\H
= gv ∈ T (E) : v ∈ SE\H .
Therefore ext(T (E)) = gv ∈ T (E) : v ∈ SE\H .
gv ) ∈ T (C ∗ (E)) : v ∈ SE\H . And
From this it follows that ext(T (C ∗ (E))) = ι−1 (
since T (C ∗ (E)) is a compact convex set we know by the Krein–Millman Theorem
that every element in T (C ∗ (E)) is a convex combination of these extreme points.
5. An example
Let E be the graph shown below:
w1
/ w2
O
/ v2
v1
w3
/ w4
= w6
{{
{
{
{{
{{
/ w5 o
w7
/ v3 .
We see that E satisfies Condition (K) since no vertex is the base point of exactly
one simple loop. Let H := {v ∈ E 0 : s(α) ≥ v for some loop α}. The set H is
hereditary and we have
H = {w5 , w6 , w7 , v3 }.
Graph traces and C ∗ -algebras
657
Let H0 = H and Hn = Hn−1 ∪ {v ∈ E 0 : 0 < |s−1 (v)| < ∞ and r(s−1 (v)) ∈
Hn−1 }. Using this definition we have H1 = H0 ∪ {w4 }. Observe that H = H0 ∪ H1
is the saturation of H. By Lemma 4.3 we know that T (E) is homeomorphic to
T (E\H); thus we need only find the extreme points of T (E\H) which is shown
here:
w1
/ w2
O
v1
w3
/ v2 .
We shall now determine ext(T (E)). We know from Theorem 4.4 that
gv1 , gv2 }.
ext(T (E)) = gv ∈ T (E) : v ∈ SE\H = {
Thus, we need only compute gv1 and gv2 . By the definition of gv1 we know that
1
n(w1 , v1 )
=
n(v1 )
2
gv1 (w1 ) =
and
gv1 (v1 ) =
1
n(v1 , v1 )
= .
n(v1 )
2
And since there are no paths from w1 , w3 , or v2 to v1 we have
gv1 (w2 ) = gv1 (w3 ) = gv1 (v2 ) = 0.
We represent gv1 here:
1
2
/0
O
/0
0
/0
1
2
0
B
/0o
0
/ 0.
In a similar way we use the definition of gv2 to conclude that
1
.
4
And since there are no paths from v1 to v2 we have gv2 (v1 ) = 0. We represent gv2
here:
gv2 (w1 ) = gv2 (w2 ) = gv2 (w3 ) = gv2 (v2 ) =
1
4
0
/
1
4O
1
4
/
1
4
/0
B0
/0o
0
/ 0.
Thus, the extreme points of E are gv1 and gv2 , which are shown above, and any
element of T (E) is a convex combination of gv1 and gv2 . Consequently, T (E) may
be viewed as a line segment between gv1 and gv2 .
Since T (E) has two extreme points we know that T (C ∗ (E)) does as well; namely
−1
gv1 ) and ι−1 (
gv2 ), where ι : T (C ∗ (E)) → T (E) is the map established in [5].
ι (
Consequently, every point τ ∈ T (C ∗ (E)) is of the form
gv1 ) + (1 − t)ι−1 (
gv2 )
τ = tι−1 (
where 0 ≤ t ≤ 1.
658
Matthew Johnson
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Department of Mathematics, University of Iowa, Iowa City, IA 52242-1419, USA
mattjohn@math.uiowa.edu
This paper is available via http://nyjm.albany.edu:8000/j/2005/11-30.html.